Question 763757
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h(t)\ =\ -16t^2\ +\ v_ot\ +\ h_o]


For *[tex \LARGE v_o\ =\ -5] ft/sec and *[tex \LARGE h_o\ =\ 300] feet


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h(t)\ =\ -16t^2\ -\ 5t\ +\ 300]


Then for *[tex \LARGE t\ =\ 2] seconds


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h(2)\ =\ -16(2)^2\ -\ 5(2)\ +\ 300]


Do the arithmetic


The ground is *[tex \LARGE h\ =\ 0] so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -16t^2\ -\ 5t\ +\ 300\ =\ 0]


Solve the quadratic for *[tex \LARGE t].


Discard the negative root.  


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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